To find the derivative of the function sinx+cosxsinx−cosx, we'll use the quotient rule. The quotient rule states that if you have a function of the form v(x)u(x), its derivative is given by:
(vu)′=v2u′v−uv′
Here, let:
- u(x)=sinx−cosx
- v(x)=sinx+cosx
Step 1: Find the derivatives of u(x) and v(x)
- u′(x)=cosx+sinx
- v′(x)=cosx−sinx
Step 2: Apply the quotient rule
Now, substitute u(x), v(x), u′(x), and v′(x) into the quotient rule formula:
dxd(sinx+cosxsinx−cosx)=(sinx+cosx)2(cosx+sinx)(sinx+cosx)−(sinx−cosx)(cosx−sinx)
Step 3: Simplify the numerator
Let's expand and simplify the numerator:
-
Expand (cosx+sinx)(sinx+cosx):
(cosx+sinx)(sinx+cosx)=cos2x+cosxsinx+sinxcosx+sin2x=cos2x+sin2x+2cosxsinx
Since cos2x+sin2x=1, the expression simplifies to:
1+2cosxsinx
-
Expand (sinx−cosx)(cosx−sinx):
(sinx−cosx)(cosx−sinx)=sin2x−sinxcosx−cosxsinx+cos2x=sin2x+cos2x−2sinxcosx
Again, sin2x+cos2x=1, so this simplifies to:
1−2sinxcosx
Thus, the numerator becomes:
(1+2cosxsinx)−(1−2sinxcosx)=4sinxcosx
Step 4: Simplify the expression
Finally, the derivative is:
(sinx+cosx)24sinxcosx
This can also be written using the identity sin2x=2sinxcosx as:
(sinx+cosx)22sin2x
Final Answer
The derivative of sinx+cosxsinx−cosx is:
(sinx+cosx)22sin2x
Would you like any further details or explanations?
Here are 5 related questions:
- What is the derivative of sinx−cosxsinx+cosx?
- How would you find the second derivative of sinx+cosxsinx−cosx?
- Can the derivative (sinx+cosx)22sin2x be simplified further?
- What is the integral of sinx+cosxsinx−cosx?
- How do you apply the quotient rule to functions involving trigonometric identities?
Tip: Remember that trigonometric identities can often simplify the differentiation or integration process, making complex expressions easier to handle.